Quadratic Equations Class 11 OP Malhotra Exe-10B ISC Maths Solutions Ch-10 Solutions. In this article you would learn about Equations Reducible to Quadratic Equations. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Quadratic Equations Class 11 OP Malhotra Exe-10B ISC Maths Solutions Ch-10

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-10	Quadratic equations
Writer	O.P. Malhotra
Exe-10(B)	Equations Reducible to Quadratic Equations.

Exercise- 10B

Quadratic Equations Class 11 OP Malhotra Exe-10B Solution.

Solve the following equations :

Que-1: x⁴ – 5x² + 6 = 0.

Sol: Given eqn. be, x4 – 5x² + 6 = 0 … (1)
put x² = t in eqn. (1) ; we have
t² – 5t + 6 = 0
⇒ (t – 2) (t – 3) = 0
⇒ t – 2 = 0 or t – 3 = 0
⇒ x² = 2 or x² = 3
⇒ x = ± √2 or x = ± √3
∴ x = ± √2, ± √3

Que-2: x⁵ + 242 = 243/x^5

Sol: Given eqn. be,
x^5 + 242 = 243/x^5 … (1)
putting x5 = y in eqn. (1) ; we have
y + 242 = (243/y) ⇒ y² + 242y – 243 = 0
⇒ y² – y + 243y – 243 = 0
⇒ y(y – 1) + 243 (y – 1) = 0
⇒ (y – 1) (y + 243) = 0
⇒ y = 1 = 0 or y + 243 = 0
⇒ y = 1 or y = – 243
⇒ x5 = 1 or x5 = (- 3)5
⇒ x = 1 or x = – 3
Thus, s = – 1, – 3

Que-3: 10x^-2 – 9 – x^-4 = 0.

Sol: Given eqn. be, 10x-2 – 9 – x-4 = 0 … (1)
putting x-2 = t in eqn. (1); we have
10t – 9 – t² = 0
⇒ t² – 10t + 9 = 0
⇒ t² – t – 9t + 9 = 0
⇒ t (t – 1) – 9 (t – 1) = 0
⇒ (t – 9)(t – 1) = 0
either t = 9 = 0 or t – 1 = 0
⇒ x-2 = 9 or x-2 – 1 = 0
⇒ 1/x² = 9 or 1/x² = 1
⇒ x = ± 1/3 or x = ± 1
Thus, x = ± 1/3, ± 1

Que-4: 3^2x – 10 x 3^x + 9 = 0.

Sol: Given eqn. be,
3^2x – 10 x 3^x + 9 = 0 …(1)
putting 3^x = t in eqn. (1) ; we have
t² – 10t + 9 = 0
⇒ t² – 9t – t + 9 = 0
⇒ t(t – 9) – 1 (t – 9) = 0
⇒ (t – 9) (t – 1) = 0
either t – 9 = 0 or t – 1 = 0
⇒ t = 9 or t = 1
⇒ 3^x = 3^x or 3^x = 3°
⇒ x = 2 or x = 0
Thus, x = 0, 2

Que-5: 2^2x-1 – 9 x 2^x-2 + 1 = 0.

Sol: Given equation be,
2^(2x-1) – 9 x 2^(x-2) + 1 = 0
⇒ (2x)² x 2^(-1) – 9 x (2^x) x 2^(-2) + 1 = 0 …(1)
putting 2x = t in eqn. (1); we have
t² x (1/2) – 9 x t x (1/4) + 1 = 0
⇒ 2t² – 9t + 4 = 0
⇒ 2t² – 8t – t + 4 = 0
⇒ 2t (t – 4) – 1 (t – 4) = 0
⇒ (2t – 1) (t – 4) = 0
either 2t – 1 = 0 or t – 4 = 0
⇒ t = (1/2) or t = 4
⇒ 2^x = (1/2) = 2^(-1) or 2^x = 2²
⇒ x = – 1 or x = 2
∴ x = – 1 or 2

Que-6: 3^2x+1 + 3² = 3^x+3 + 3^x.

Sol: Given eqn. be,
3^2x+1 + 3² = 3^x+3 + 3^x
⇒ (3^x)² x 3 + 9 = 3^x (3³ + 1) …(1)
putting 3^x = t in eqn. (1) ; we have
3t² + 9 = 28t ⇒ 3t² – 28t + 9 = 0
⇒ 3t² – 27t – t + 9 = 0
⇒ 3t(t – 9) – 1 (t – 9) = 0
⇒ (t – 9) (3t – 1) = 0
either t – 9 = 0 or 3t – 1 = 0
⇒ t = 9 or t = (1/3)
⇒ 3^x = 3² or 3^x = 3^-1
⇒ x = 2 or x = – 1
Thus, x = 2, – 1

Que-7: √(x²-3x) = 4x²−12x−3

Sol: Given equation be,
√(x²-3x) = 4x²−12x−3 – 3 …(1)
putting x² – 3x = t in eqn. (1); we have
√t = 4t – 3 ;
on squaring both sides ; we have
⇒ t = (4t – 3)² ⇒ t = 16t² – 24t + 9 ⇒ 16t² – 25t + 9 = 0

Que-8: √{(x²+2)/(x²-2)} + 6√{(x²-2)/(x²+2)} = 5

Sol: Given equation be
√{(x²+2)/(x²-2)} + 6√{(x²-2)/(x²+2)} = 5
Putting √{(x²+2)/(x²-2)} = y in eqn. (1) ; we have
y + (6/y) = 5
⇒ y² – 5y + 6 = 0
⇒ (y – 2)(y – 3) = 0
⇒ y – 2 = 0 or y – 3 = 0
⇒ y = 2 or y = 3
Case-I: When y = 2
⇒ √{(x²+2)/(x²-2)} = 2
On squaring both sides ; we have
{(x²+2)/(x²-2)} = 4
⇒ x² + 2 = 4x² – 8
⇒ 3x² = 10
⇒ x² = 10/3
x = ±√(10/3)
When y = 3 ⇒ √{(x²+2)/(x²-2)} = 3
On squaring both sides ; we have
{(x²+2)/(x²-2)} = 9 ⇒ x² + 2 = 9x² – 18
⇒ 8x² = 20
⇒ x² = 52
⇒ x = ± √5/2
Hence x = ± √10/3,  ±√5/2

Que-9: √{(2x²+1)/(x²-1)} + 6 √{(x²-1)/(2x²+1)} = 5.

Sol: Given equation be
√{(2x²+1)/(x²-1)} + 6 √{(x²-1)/(2x²+1)} = 5.   ….. (1)
Putting √{(2x²+1)/(x²-1)} = t in eqn (1) we have,
t + (6/t) = 5
⇒ t² – 5t + 6 = 0
⇒ (t – 2)(t – 3) = 0
either t – 2 = 0 or t – 3 = 0
⇒ t = 2 or t = 3
Case-I: When t = 2
√{(2x²+1)/(x²-1)} = 2
On squaring both sides ; we have 2*2 + 1
{(2x²+1)/(x²-1)} = 4
⇒ 2x² + 1 = 4x² – 4
⇒ 2x² = 5
⇒ x = ±√5/2
Case-II: When t = 3
√{(2x²+1)/(x²-1)} = 3
On squaring both sides ; we have
{(2x²+1)/(x²-1)} = 9
2x²+1 = 9x²-9
7x² = 10
x² = 10/7
x = ±√10/7
x = ±√5/2,  ±√10/7

Que-10: x (x – 1) (x + 2) (x- 3) + 8 = 0.

Sol: Given equation be
x (x – 1) (x + 2) (x – 3) + 8 = 0
⇒ (x² – x) [x² – x – 6] + 8 = 0 … (1)
putting x² – x = t in eqn. (1) ; we have
t(t – 6) + 8 = 0 ⇒ t² – 6t + 8 = 0
⇒ t² – 4t – 2t + 8 = 0
⇒ t(t – 4) – 2(t – 4) = 0
⇒ (t – 4)(t – 2) = 0
either t – 4 = 0 or t – 2 = 0
⇒ t = 4 or t = 2
Case-I: When t – 4
⇒ x² – x – 4 = 0
⇒ x = [−b±√(b²-4ac)]/2a
⇒ x = [−(−1)±√{(-1)²-4×1×(-4)}/2
⇒ x = (1±√17)/2
Case-II: When t = 2 ⇒ x² – x – 2 = 0
⇒ x² – 2x + x – 2 = 0
⇒ x (x – 2) + 1 (x – 2) = 0
⇒ (x + 1) (x – 2) = 0
⇒ x = – 1, 2
Hence x = – 1, 2, {(1±√17)/2}

Que-11: (x – 7) (x – 3) (x + 1) (x + 5) = 1680.

Sol: Given eqn. be
(x – 7) (x – 3) (x + 1)(x + 5) = 1680
⇒ {(x – 7) (x + 5)} {(x-3) (x+ 1)} = 1680
⇒ {x² – 2x – 35} {x² – 2x – 3} = 1680 … (1)
putting x² – 2x = t in eqn. (1); we get
(t – 35) (t – 3) = 1680
⇒ t² – 38t + 105 – 1680 = 0
⇒ t² – 38t – 1575 = 0
∴ t = [38±√(1444+6300)]/2 = (38±88)/2
⇒ t = 63, – 25
Case-I: When t = 63 ⇒ x² – 2x – 63 = 0

Que-12: (2x – 7) (x² – 9) (2x + 5) = 91.

Sol: Given equation be,
(2x – 7) (x² – 9) (2x + 5) = 91
⇒ {(2x – 7) (x + 3)} {(x – 3)(2x + 5)} = 91
⇒ (2x² – x – 21)(2x² – x – 15) = 91 …(1)
putting 2x² – x = t in eqn. (1) ; we have
(t – 21) (t – 15) = 91
⇒ t² – 36t + 224 = 0
⇒ t² – 8t – 28t + 224 = 0
⇒ t (t – 8) – 28 (t – 8) = 0
⇒ (t – 8) (t – 28) = 0
either t – 8 = 0 or t – 28 = 0
⇒ t = 8 or t = 28
Case-I: When t = 8 ⇒ 2x² – x – 8 = 0
⇒ x = [−(−1)±√{(−1)²−4×2×(−8)}]/(2×2)
⇒ x = (1±√65)/4
Case-II: When t = 28
⇒ 2x² – x – 28 = 0
⇒ 2x² – 8x + 7x – 28 = 0
⇒ 2x (x – 4) + 7 (x – 4) = 0
⇒ (x – 4) (2x + 7) = 0
either x – 4 = 0 or 2x + 7 = 0
⇒ x = 4 or x = – 7/2
Hence, x = + 4, (–7/2),  {(1±√65)/4}

Que-13: By substituting y = 2^x, or otherwise, solve the equation
2^2x + 2^x+2 – 4 x 2³ = 0.

Sol: Given eqn. be,
2^2x + 2^x+2 – 4 x 2³ = 0 …(1)
putting 2^2x = y in eqn. (1) ; we have
y² + y . 2² – 32 = 0
⇒ y² + 4y – 32 = 0
⇒ y² + 8y – 4y – 32 = 0
⇒ y(y + 8) – 4(y + 8) = 0
⇒ (y + 8)(y – 4) = 0
either y + 8 = 0 or y – 4 = 0
⇒ y = – 8 or y = 4
⇒ 2^x = – 8
it has no real solution (∵ 2^x > 0)
or 2^x = 4 = 2² ∴ x = 2
Thus x = 2

Que-14: 2^x² : 2^x = 8 : 1.

Sol: Given : 2^x² : 2^x = 8 : 1
2^x² / 2^x = 8
2^x² = 8.2^x

Que-15: 2^2x+3 + 2^x+3 = 1 + 2^x

Sol: Given eqn. be 2^2x+3 + 2^x+3 = 1 + 2^x
⇒ (2^x)² x 2³ + 2^x x 2³ = 1 + 2^x …(1)
putting 2^x = t in eqn. (1) ;
we have 8t² + 8t = 1 + t
⇒ 8t² + 7t – 1 = 0
⇒ 8t² + 8t – t – 1 = 0
⇒ 8t (t + 1) – 1 (t + 1) = 0
⇒ (t + 1) (8t – 1) = 0
either t + 1 = 0 or 8t – 1 = 0
⇒ t = – 1 or t = 1/8
⇒ 2^x = – 1, which is not possible as 2^x > 0
when t = 1/8
=  2^x = 2^-3 ⇒ x = – 3

Que-16: (4^x) − 3^{x−(1/2)} = 3^{x+(1/2)} − 2^(2x−1)

Sol: Given equation be
(4^x) − 3^{x−(1/2)} = 3^{x+(1/2)} − 2^(2x−1)

–: End of Quadratic Equations Class 11 OP Malhotra Exe-10B ISC Maths Ch-10 Solutions. :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions

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